What are the divisors of 1983?

1, 3, 661, 1983

There is a total of 4 positive divisors.
The sum of these divisors is 2648.
The arithmetic mean is 662.

4 odd divisors

1, 3, 661, 1983

How to compute the divisors of 1983?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 1983 by each of the numbers from 1 to 1983 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

1983 / 1 = 1983 (the remainder is 0, so 1 is a divisor of 1983)
1983 / 2 = 991.5 (the remainder is 1, so 2 is not a divisor of 1983)
1983 / 3 = 661 (the remainder is 0, so 3 is a divisor of 1983)
...
1983 / 1982 = 1.0005045408678 (the remainder is 1, so 1982 is not a divisor of 1983)
1983 / 1983 = 1 (the remainder is 0, so 1983 is a divisor of 1983)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 1983 (i.e. 44.530888156425). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

1983 / 1 = 1983 (the remainder is 0, so 1 and 1983 are divisors of 1983)
1983 / 2 = 991.5 (the remainder is 1, so 2 is not a divisor of 1983)
1983 / 3 = 661 (the remainder is 0, so 3 and 661 are divisors of 1983)
...
1983 / 43 = 46.116279069767 (the remainder is 5, so 43 is not a divisor of 1983)
1983 / 44 = 45.068181818182 (the remainder is 3, so 44 is not a divisor of 1983)